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$a+b+c=12$ and $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$. Find $a^3 + b^3 +c^3 = ?$

My approach is to use AM-GM inequality. Is it correct?

amWhy
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  • Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational? – A. Pongrácz Dec 19 '18 at 15:21

1 Answers1

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If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.

Namely, consider the following 12 numbers:

$\frac{a}{3}, \frac{a}{3}, \frac{a}{3}, \frac{b}{4}, \frac{b}{4}, \frac{b}{4}, \frac{b}{4}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}$.

The AM is $1$, as $a+b+c=12$.

Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields $\frac{a^3b^4c^5}{3^3\cdot 4^4\cdot 5^5}\leq 1$. As $a^3b^4c^5 = 0.1\cdot 600^3 = 3^3\cdot 4^4\cdot 5^5$, the above inequality is sharp: it holds with equality. Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal. Hence, $\frac{a}{3}= \frac{b}{4}= \frac{c}{5}$. Using $a+b+c=12$ we obtain $a=3, b=4, c=5$. Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.

A. Pongrácz
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